Need Help - dragging a trunk up a loading ramp

[Lolagoeslala]

Need Help -- dragging a trunk up a loading ramp

Homework Statement

A man is dragging a 68 kg trunk up the loading ramp of a movers truck, the ramp has a slope of 20deg and the man pulls upward with a force F whose direction makes an angle of 30deg with the ramp. The coefficient of kinetic friction between trunk and the ramp is 0.28. What force F is necessary to accelerate the trunk up the loading ramp at a rate of 0.55m/s^2?

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Homework Equations

The Attempt at a Solution

I broke the F in two parts one on the verticle and the other on the horizontal same with the gravity but now i am confused...

 

[TSny]

Find the components of the forces that are either parallel to the ramp or perpendicular to the ramp (rather than horizontal and vertical components.) The reason is that if you know the direction of the acceleration, then it's almost always a good idea to work with the components that are in the direction of the acceleration or perpendicular to the acceleration. That will make application of the 2nd law easier. In this problem you know the direction of the acceleration is up the ramp.

 

[Lolagoeslala]

Find the components of the forces that are either parallel to the ramp or perpendicular to the ramp (rather than horizontal and vertical components.) The reason is that if you know the direction of the acceleration, then it's almost always a good idea to work with the components that are in the direction of the acceleration or perpendicular to the acceleration. That will make application of the 2nd law easier. In this problem you know the direction of the acceleration is up the ramp.

Hey i did do that..
then i got this answer...
but the checking was not correct..
My solution:
I broke the gravity components into perpendicular and parallel so
as a verticle equilibrium i got:
mgcos20deg=Fn+Fsin30deg

and as horizontal equilibrium i got:
Fcos30deg=μFn+mgsin20deg

I rearranged the horizontal equilibrium for F to find the Fn in the first equation
like the following:
F= ((μFn+mgsin20deg)/cos30deg)
then:

mgcos20deg=Fn+((μFn+mgsin20deg)/cos30deg)(sin30deg)
By adding into all the given information i got 425.81N=Fn

I plugged that into the friction formula Ff=μFn to find friction which i got was 119.2268N

then i used this equation to use the acceleration given and find the F:
Fnet=Fcos30deg-mgsin20deg-Ff
ma=Fcos30deg-mgsin20deg-Ff
F= 444.04 N

But when i am checking the answer in the verticle equilibrium from before:
mgcos20deg=Fn+Fsin30deg
626.21N=425.8N+222.02N
626.21N=647.82N

 

[TSny]

OK, so you are using "horizontal" to mean parallel to the ramp, and "vertical" to mean perpendicular to the ramp.

I broke the gravity components into perpendicular and parallel so
as a verticle equilibrium i got:
mgcos20deg=Fn+Fsin30deg

Check to see if you really want to use 20 degrees here. [Edit: MY fault. 20 degrees is correct!]

and as horizontal equilibrium i got:
Fcos30deg=μFn+mgsin20deg

Are you dealing with an equilibrium situation in the direction parallel to the ramp?

 

[Lolagoeslala]

OK, so you are using "horizontal" to mean parallel to the ramp, and "vertical" to mean perpendicular to the ramp.

Check to see if you really want to use 20 degrees here.

Are you dealing with an equilibrium situation in the direction parallel to the ramp?

Well yes i think you would use 20 degrees because that is the gravity angle.
Yes i am using the direction parallel to the ramp...

 

[TSny]

What angle does the force of gravity make with respect to the direction of the ramp?

 

[TSny]

Never mind, you are correct! The angle you want is 20 deg. Sorry.

 

[Lolagoeslala]

Never mind, you are correct! Sorry.

But how come when i do the checking the answer is totally different? Which other way would you recommend to do this question?

 

[TSny]

How are you going to include the fact that the trunk is accelerating?

 

[Lolagoeslala]

But how are you going to include the fact that the trunk is accelerating?

Well see i used the this equation
Fnet=Fcos30deg-mgsin20deg-Ff
which looks at the horizontal (parallel to ramp) forces, since the verticle forces cancel out...
To find the F which includes the a= acceleration of 0.55m/s^2 and then i did the following math...
ma=Fcos30deg-mgsin20deg-Ff
F= 444.04 N
And that was the force i got...

 

[TSny]

Go back to where you said:

and as horizontal equilibrium i got:
Fcos30deg=μFn+mgsin20deg

This is not correct, the forces are not in equilibrium in the horizontal direction.

 

[Lolagoeslala]

Go back to where you said:

This is not correct, the forces are not in equilibrium in the horizontal direction.

Why are they not equilibrium... i mean when you create the FBD,,,, the gravity and the friction force pull down while the applied force (F) is pulling upwards with the acceleration of 0.55m/s^2

 

[TSny]

If the force components parallel to the ramp were in equilibrium, the trunk would not accelerate up the ramp. You essentially have two unknowns: Fn and F. Need two equations. One equation is your equation for the components of the forces perpendicular to the ramp. For the other, use your equation ma = Fcos30 - mgsin20 - Ff.

 

[Lolagoeslala]

If the force components parallel to the ramp were in equilibrium, the trunk would not accelerate up the ramp. You essentially have two unknowns: Fn and F. Need two equations. One equation is your equation for the components of the forces perpendicular to the ramp. For the other, use your equation ma = Fcos30 - mgsin20 - Ff.

Umm so you want me to add these two equation,,
i mean the following:

mgcos20deg=Fn+Fsin30deg
and ma = Fcos30 - mgsin20 -Ff...
but i am not sure what to do...
i mean could u show me how o get started i understand your appoint on the equilibrium,,,

 

[TSny]

In the second equation you know how you could express Ff in terms of Fn and the coefficient of friction. So, you then have two equations with two unknowns (Fn and F). See if you can eliminate Fn between the equations to get a single equation with just the unknown F.

 

[Lolagoeslala]

In the second equation you know how you could express Ff in terms of Fn and the coefficient of friction. So, you then have two equations with two unknowns (Fn and F). See if you can eliminate Fn between the equations to get a single equation with just the unknown F.

mgcos20deg=Fn+Fsin30deg
ma = Fcos30 - mgsin20 -μFN

Umm how about i do this...

F= (ma+μFn+mgsin20)/cos30

then i plug into the other equation as mgcos20deg=Fn +((ma+μFn+mgsin20)/cos30)(sin30)
Find the Fn and then use that to find the F?

 

[TSny]

That would work. But it might be less work if you solve the first equation for Fn first. Then substitute for Fn in the second equation.

 

[TSny]

Or, what would happen if you multiplied the first equation through by μ and then add the two equations?

 

[Lolagoeslala]

Or, what would happen if you multiplied the first equation through by μ and then add the two equations?

oh that would work...
let me try it right away..

however we can't just solve the first equation because f is missing..

 

[Lolagoeslala]

Or, what would happen if you multiplied the first equation through by μ and then add the two equations?

Omg i found the Fn and the F and guess what the checking seems to work!

 

[TSny]

Great! Excellent work.